What is the limit as x approaches 0 of (2x^2+x)/(5x) ?

The limit as x approaches 0 of (2x^2+x)/(5x) is 1/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 0 of (2x^2+x)/(5x)

x \to 0 lim (\frac{2 x ^{2} + x}{5 x})

\frac{1}{5}

Decimal Notation

0.2

Solution steps

x \to 0 lim (\frac{2 x ^{2} + x}{5 x})

Simplify

\frac{2 x ^{2} + x}{5 x} : \frac{1}{5} (2 x + 1)

= x \to 0 lim (\frac{1}{5} (2 x + 1))

Plug in the value

x = 0

= \frac{1}{5} (2 \cdot 0 + 1)

Simplify

= \frac{1}{5}

x \to 2 lim (\frac{( x - 2 )}{x - 2})

x \to 0 + lim (5 x ln (x))

x \to 0 lim (\frac{sin ( ω ) x}{x})

x \to 0 lim (- 1 + x)

x \to 0 lim (\frac{\frac{4}{x + 2} - 2}{x})

What is the limit as x approaches 0 of (2x^2+x)/(5x) ?
The limit as x approaches 0 of (2x^2+x)/(5x) is 1/5